2D Force Problems

Question

3. A 50 N applied force (pulling 30° up from the horizontal) accelerates a 5 kg box across a horizontal floor. a) What is the normal force on the box?



b) If the coefficient of friction between the floor and the box is 0.15, what is the acceleration of the box?

Answer 1

F_n = normal force

 

mg = weight of box, m = 5 kg

 

F = force (applied at angle θ = 30º above horizontal) = 50 N

 

(a)

 

Since there is no upward acceleration in this problem, the vertical forces must be in equilibrium.

 

F_n + F sinθ = mg

 

F_n = mg - F sinθ = (5 kg)(9.81 m/s²) - (50 N)sin30º = 24.05 N

 

(b)

 

In the horizontal direction,

 

F cosθ - μF_n = ma

 

a = (F cosθ - μF_n)/m = (50cos30º - 0.15*24.05)/5 m/s² = 7.94 m/s²